Giải các bất phương trình :
$\begin{array}{l}
1)\,\,\,\,\,\frac{{{{\log }_5}{{\left( {{x^2} - 4x - 11} \right)}^2} - {{\log }_{11}}{{\left( {{x^2} - 4x - 11} \right)}^3}}}{{2 - 5x - 3{x^2}}} \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
2)\,\,\,\,\frac{{{{\log }_2}{{\left( {{x^2} - 2x - 7} \right)}^5} - {{\log }_3}{{\left( {{x^2} - 2x - 7} \right)}^8}}}{{3{x^2} - 13x + 4}} \le 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{array}$
1) Điều kiện: $\left\{ \begin{array}{l}x^2-4x-11>0\\2-5x-3x^2\neq0 \end{array} \right.$
Khi đó ta có:
$(1) \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 4x - 11 > 0\\
2 - 5x - 3{x^2} \ne 0\\
\frac{{{{\log }_5}\left( {{x^2} - 4x - 11} \right)}}{{3{x^2} + 5x - 2}}\left( {2 - \frac{3}{{{{\log }_5}11}}} \right) \le 0\,\,\,\,\,\,\,\,\,\,\,(*)
\end{array} \right.$
$
Do  2 - \frac{3}{{{{\log }_5}11}} < 0 $  nên $(*)  \Leftrightarrow \left\{ \begin{array}{l}
3{x^2} + 5x - 2 > 0\\
{x^2} - 4x - 11 \ge 1
\end{array} \right.$
                                               Hoặc $ \Leftrightarrow \left\{ \begin{array}{l}
3{x^2} + 5x - 2 < 0\\
{x^2} - 4x - 11 \le 1
\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} (3x-1)(x+2)>0\\(x-6)(x+2)\geq 0 \end{array} \right.\\\left\{ \begin{array}{l}(3x-1)(x+2)<0\\(x-6)(x+2)\leq 0 \end{array} \right.\end{array} \right.\Leftrightarrow \left[\begin{array}{l} x\geq 6\\ x<-2\\-2<x<\frac{1}{3} \end{array} \right.$.
Kết hợp với điều kiện, vậy PT đã cho có nghiệm:
           $x <  - 2;\,\, - 2 < x < 2 - \sqrt {15} ;\,\,x \ge 6$

$2)$ 
Điều kiện: $\left\{ \begin{array}{l}x^2-2x-7>0\\3x^2-13x+4\neq0 \end{array} \right.$
Khi đó ta có:
$(2)\Leftrightarrow \frac{5.\log_23.\log_3(x^2-2x-7)-8\log_3(x^2-2x-7)}{3x^2-13x+4}\leq 0$
$\Leftrightarrow \frac{\log_3(x^2-2x-7)(5\log_23-8)}{3x^2-13x+4}\leq 0$
$\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} x^2-2x-7\geq 1\\3x^2-13x+4>0 \end{array} \right.\\\left\{ \begin{array}{l} x^2-2x-7\leq 1\\3x^2-13x+4<0 \end{array} \right. \end{array} \right.(do  5\log_23-8<0)$
$\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} (x-4)(x+2)\geq 0\\(3x-1)(x-4)>0 \end{array} \right.\\\left\{ \begin{array}{l}(x-4)(x+2)\leq 0\\(3x-1)(x-4)<0 \end{array} \right. \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x>4\\\frac{1}{3}<x<4\\x\leq -2 \end{array} \right.$
Kết hợp với điều kiện, vậy PT đã cho có nghiệm:
            $x \le  - 2;\,\,1 + \sqrt 8  < x < 4;\,\,x > 4$

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